A CONTINUED FRACTION EXPANSION FOR A q-tangent FUNCTION

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Sémiaire Lotharigie de Combiatoire 45 001, Article B45b A CONTINUED FRACTION EXPANSION FOR A q-tangent FUNCTION MARKUS FULMEK Abstract. We prove a cotiued fractio expasio for a certai q taget fuctio that was cojectured by Prodiger. 1. Itroductio I [4], Prodiger defied the followig q trigoometric fuctios 1 +1 si q z [ + 1] q! q, cos q z 0 0 Here, we use stadard q otatio: 1 q. [] q! [] q : 1 q 1 q, [] q! : [1] q [] q... [] q, a; q : 1 a1 aq... 1 aq 1. These q fuctios are variatios of Jackso s [] q-sie ad q-cosie fuctios. For the q taget fuctio ta q siq cos q, Prodiger cojectured the followig cotiued fractio expasio see [4, Cojecture 10]: z ta q z [1] q q 0 [3] q q [5] q q 1 [7] q q 9. 1 Here, the powers of q are of the form 1 1 1/ + 1. The purpose of this ote is to prove this statemet. I our proof, we make use of the polyomials see [3,, 11] A z ad B z, which Date: Jauary 5, 001.

MARKUS FULMEK are give recursively by A z b A 1 z A z, B z b B 1 z B z; 3 with iitial coditios see [3,, 1] A 1 1, B 1 0, A 0 b 0, B 0 1, where b 0 0, b [ 1] q q 1 1 1/ +1. As is well kow see [3, ], the cotiued fractio termiated after the term b is equal to A B, whece 1 follows from the assertio A cos q +zb si q O+1, 4 i.e., the leadig coefficiets of z vaish i 4. I Sectio we give a proof of 4 ad thus of 1.. The proof Both A ad B are polyomials i : A z c,j j, B z j j d,j j. Observe that from the recursios ad 3 we obtai immediately c,k b c 1,k c,k 1 ad d,k b d 1,k d,k 1, 5 with iitial coditios c 0,k d 0,k c 1,k d 1,k c,0 d, 1 0, c 1,1 1, d 0,0 1. Give this otatio, we have to prove the followig assertio for the coefficiets of k i 4: For 1, 0 k, there holds k 1 k i k 1 k i 1 c,i [k i] q! qk i + d,i [k i 1] q! qk i 1 0. 6 I fact, we shall state ad prove a slightly more geeral assertio: Lemma 1. Give the above defiitios, we have for all 1, k 0: k 1 1 i q k i 1 [k i ] q! c,i+1 + d,i [k i 1] q 1 q 5+3 1 1k 4 1 k+8k +8 8k+4 1 /8 k sk [s] q. 7 [k] q!

A CONTINUED FRACTION EXPANSION 3 Note that the left had side of 7 is the same as i 6, ad the right had side of 7 vaishes for 0 k. Hece 6 ad thus Prodiger s cojecture is a immediate cosequece of Lemma 1. Proof. We perform a iductio o k for arbitrary. The case k 0 is immediate. For the case k 1, observe that c,1 d,0 b. For the iductive step k 1 k, we shall rewrite the recursios 5 i the followig way: c,k c i,k 1 s1 b j, d,k d i,k 1 b j Substitutio of these recursios ito 7 ad iterchage of summatios trasform the idetity ito q k 1 1 [k 1] q s1 b s [k 1] q! rhsi, k 1 + b j rhs, k, where rhs, k deotes the right had side of 7. Now we use the iductio hypothesis. As it turs out, factorizatio of powers of q from rhsi, k 1 b j yields the same power for i ad i + 1, whece we ca group these terms together. After several steps of simplificatio we arrive at the followig idetity: 4 q k+6 ; q 4 j q k+4 ; q 4 j q 17/ k ; q 4 j q 17/ k ; q 4 j q k 1j q 7 ; q 4 j q 9 ; q 4 j q 9/ k ; q 4 j q 9/ k ; q 4 j q; q q1 q k 1 1 q k 1 q 9 k 1 q1 q 3 1 q 5. 1 q k 1 q 3 ; q 1 1 q 1+ 1 +k 1 k+ 4k+4 /4 q k ; q + q; q + χ1 q k 1 q k+1/ q k + ; q 1 0, 8 where χ 1 for eve ad 0 for odd.

4 MARKUS FULMEK The sum ca be evaluated by meas of the very well poised 6 φ 5 summatio formula [1,.7.1; Appedix II.0]: a; q j aq; q j aq; q j b; q j c; q j d; q j q; q j a; q j a; q j aq ; q b j aq ; q c j aq ; q d j aq j bcd aq; q aq ; q bc aq ; q bd aq ; q cd aq ; q b aq ; q c aq ; q d aq ; q. 9 bcd The sum we are actually iterested i does ot exted to ifiity, so we rewrite is as follows: 4 s, k, j s, k, j j s, k, j s, k, j s, k, s, k, j + s, k,, where s, k, j deotes the summad i 8. Now, replacig q by q 4, a by q k+8a+9, b by q k+4a+4, c by q k+4a+6 ad d by q 4 i the summad of 9 gives s,k,j+a times the fractio q k+8a+9 ;q 4 j q 4 ;q 4 j s,k,a q k+8a+9 ;q 4 j q 4 ;q 4 j, which cacels. So we obtai after some simplificatio: x q k+6 ; q 4 j q k+4 ; q 4 j q 17/ k ; q 4 j q 17/ k ; q 4 j q 7 ; q 4 j q 9 ; q 4 j q 9/ k ; q 4 j q 9/ k ; q 4 j q k 1j 1 q 3 1 q 5 1 q k 1 1 q 9 k 1 q k 1x+1 q k+4 ; q x+. q 3 ; q x+ Substitutio of this evaluatio i 8 ad simplificatio yield for both cases eve N ad odd N 1 the same equatio q k+ ; q N 1 q k NN 1 q k 4N+ ; q N 1, which, of course, is true. This fiishes the proof. Refereces [1] G. Gasper ad M. Rahma, Basic hypergeometric series, Ecyclopedia of Mathematics ad its Applicatios 35, Cambridge Uiversity Press, Cambridge, 1990. [] F.H. Jackso, A basic sie ad cosie with symbolic solutios of certai differetial equatios, Proc. Ediburgh Math. Soc.,, 1904, 8 39. [3] O. Perro, Die Lehre vo de Kettebrüche, 1. Bad, B.G. Teuber, Stuttgart, 1977. [4] H. Prodiger, Combiatorics of geometrically distributed radom variables: New q-taget a q-secat umbers, It. J. Math. Math. Sci. to appear.

A CONTINUED FRACTION EXPANSION 5 Istitut für Mathematik der Uiversität Wie, Strudlhofgasse 4, A-1090 Wie, Austria E-mail address: Markus.Fulmek@Uivie.Ac.At WWW: http://www.mat.uivie.ac.at/~mfulmek

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